How do you integrate #int (2x +1) / ((x - 2)(x^2 + 1)) # using partial fractions?

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How do you integrate #int (2x +1) / ((x - 2)(x^2 + 1)) # using partial fractions?

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Answer 1 · 2016-04-01T03:49:24

#int(2x+1)/((x-2)(x^2+1))dx=ln|x-2|-1/2ln(x^2+1)+c#

Explanation:

#int(2x+1)/((x-2)(x^2+1))dx=intA/(x-2)dx+int(Bx+C)/(x^2+1)dx#
#0x^2+2x+1=A(x^2+1)+(Bx+C)(x-2)#
Equating coefficients:
#A+B=0#
#C-2B=2#
#A-2C=1#
Solving simultaneously yields #A=1#, #B=-1# and #C=0#
#=int1/(x-2)dx-intx/(x^2+1)dx#
#=ln|x-2|-1/2ln(x^2+1)+c#

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How do you integrate #int (2x +1) / ((x - 2)(x^2 + 1)) # using partial fractions?

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